**Equation of a Slanted Ellipse:**The equation \(x^2 + xy + y^2 = 9\) represents a slanted ellipse shown in the figure below:**Graph Description:**The graph illustrates an ellipse centered around the origin. It is positioned at an angle, creating a slanted appearance relative to the coordinate axes. The ellipse extends between approximately -3 to 3 along the x-axis and -3 to 3 along the y-axis.**Problem Solving:**1. Think of \(y\) as a function of \(x\). Differentiating implicitly and solving for \(y'\) gives: \[ y' = \_\_\_\_\_ \] *(Your answer will depend on \(x\) and \(y\).)*2. The ellipse has two horizontal tangents. The upper one has the equation: \[ y = \_\_\_\_\_ \]3. The rightmost vertical tangent has the equation: \[ x = \_\_\_\_\_ \] This tangent touches the ellipse where: \[ y = \_\_\_\_\_ \]**Hint:**- A horizontal tangent is characterized by \(y' = 0\). To find the vertical tangent, utilize symmetry or consider \(x\) as a function of \(y\). Differentiate implicitly, solve for \(x'\), and set \(x' = 0\).

Given that x2+xy+y2=9 Differentiate w.r.t x , 2x+y+xy'+2yy'=0 y'=-2x+yx+2y Now, differentiate w.r.t. y , 2xx'+x+yx'+2y=0 x'=-x+2y2x+yFor horizontal tangent , y'=0-2x+yx+2y=0x=-y2 Substitute this in main equation , x2+xy+y2=9-y22+-y2y+y2=9y2-2y2+4y2=363y2=36y=±12 For vertical tangent , x'=0-x+2y2x+y=0y=-x2 Substitute this in main equation , x2+xy+y2=9x2+-x2x+-x22=9x2-2x2+4x2=363x2=36x=±12 CONCLUSION: Right most equation for x of the tangent will be, x=+12 Left most equation for x of the tangent will be , x=-12 Right most equation for y of the tangent will be , y=-x2=-122=-3